I do not think that you have written what you meant to write.
If n is a natural number [n]=n. Is that what you meant.
Or did you mean something like ?
Hello. I need to find the supremum and infimum of the set S={sqrt(n) - [n] : n in N the natural numbers}. [n] stands for the greatest integer m in Z such that m is less than or equal to n.
I assume the infimum is zero since you can't take the square root of a negative number (and fractions between 0 and 1 will be 0 in [n]) and I'm pretty sure there is no supremum since the function is unbounded. How would I prove these (both supremum and infimum) for an epsilon greater than zero?
and
i'm not sure what you are talking about with the epsilon greater than zero thing. that is not something we use for supremums and infimums, at least not that i've seen. are you sure you're not talking about some limit here?
i agree with what Plato said. something is off here. (which means my answer may be wrong, i think. make sure you have the right question so i can modify it)
It is not at all clear that you are ready for this problem!
First of all, no one simply picked values of n.
I showed that a lower bound belonged to the set.
In that case that value is of the set.
However, you will not ever find a value of n for which there is an upper bound that belongs to the set.
You must show that 1 is an upper bound and is also the least upper bound.
OK, I guess I see what you're doing. You showed that 0 is a lower bound and 1 is an upper bound. I guess I don't know how to show that 1 is a part of the set since you can't pick values that make the function equal one. You can for zero, but you can't for one. That's correct, right? So obviously this is a little bit tougher to show and I don't know how to.
It seems completely different from the first one (the lower bound) since one isn't part of the set; it's just an upper bound. How do I prove that it's the lowest upper bound?